$ D = \left[\begin{array}{rrr}1 & -2 & 4 \\ 3 & -1 & 4\end{array}\right]$ $ B = \left[\begin{array}{rr}-1 & 5 \\ 2 & -2 \\ 2 & 1\end{array}\right]$ What is $ D B$ ?
Solution: Because $ D$ has dimensions $(2\times3)$ and $ B$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D B = \left[\begin{array}{rrr}{1} & {-2} & {4} \\ {3} & {-1} & {4}\end{array}\right] \left[\begin{array}{rr}{-1} & \color{#DF0030}{5} \\ {2} & \color{#DF0030}{-2} \\ {2} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ B$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ B$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ B$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{-1}+{-2}\cdot{2}+{4}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{-1}+{-2}\cdot{2}+{4}\cdot{2} & ? \\ {3}\cdot{-1}+{-1}\cdot{2}+{4}\cdot{2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{-1}+{-2}\cdot{2}+{4}\cdot{2} & {1}\cdot\color{#DF0030}{5}+{-2}\cdot\color{#DF0030}{-2}+{4}\cdot\color{#DF0030}{1} \\ {3}\cdot{-1}+{-1}\cdot{2}+{4}\cdot{2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{-1}+{-2}\cdot{2}+{4}\cdot{2} & {1}\cdot\color{#DF0030}{5}+{-2}\cdot\color{#DF0030}{-2}+{4}\cdot\color{#DF0030}{1} \\ {3}\cdot{-1}+{-1}\cdot{2}+{4}\cdot{2} & {3}\cdot\color{#DF0030}{5}+{-1}\cdot\color{#DF0030}{-2}+{4}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}3 & 13 \\ 3 & 21\end{array}\right] $